Find $\dfrac{d}{dx}\left[ x^4\sin(x)\cos(x) \right]$. Choose 1 answer: Choose 1 answer: (Choice A) A $-4x^3\cos(x)\sin(x)$ (Choice B) B $4x^3\sin(x)\cos(x)+x^4\cos^2(x)-x^4\sin^2(x)$ (Choice C) C $4x^3\sin(x)\cos(x)+x^4\cos^2(x)$ (Choice D) D $-4\sin^3(\cos(x))\cos(\cos(x))\sin(x)$
$x^4\sin(x)\cos(x)$ is a product of three functions. Let... $u(x)=x^4$ $v(x)=\sin(x)$ $w(x)=\cos(x)$... then $x^4\sin(x)\cos(x)=u(x)\cdot v(x) \cdot w(x)$. To find $\dfrac{d}{dx}\left[ x^4\sin(x)\cos(x) \right]$, we will need to use the product rule twice! $\begin{aligned} &\phantom{=}\dfrac{d}{dx}\left[u(x)\cdot v(x) \cdot w(x)\right] \\\\ &=u'(x)\cdot \Bigl(v(x)\cdot w(x) \Bigr)+u(x)\cdot\dfrac{d}{dx}\left[v(x)\cdot w(x)\right]&\gray{\text{Product rule}} \\\\ &=u'(x)\cdot \Bigl(v(x)\cdot w(x) \Bigr)+u(x)\cdot\Bigl(v'(x)\cdot w(x)+v(x)w'(x)\Bigr)&\gray{\text{Product rule}} \\\\ &= u'(x)\cdot v(x) \cdot w(x) + u(x) \cdot v'(x) \cdot w(x) + u(x) \cdot v(x) \cdot w'(x) \end{aligned}$ Let's differentiate $u$, $v$, and $w$ : $u'(x)=4x^3$ $v'(x)=\cos(x)$ $w'(x)=-\sin(x)$ Now we can plug the equations for $u$, $v$, $w$, $u'$, $v'$, AND $w'$ into the expression we got: $\begin{aligned} &\phantom{=}u'(x)\cdot v(x) \cdot w(x) + u(x) \cdot v'(x) \cdot w(x) + u(x) \cdot v(x) \cdot w'(x) \\\\ &=4x^3 \cdot \sin(x) \cdot \cos(x) + x^4 \cdot \cos(x) \cdot \cos(x) + x^4 \cdot \sin(x) \cdot -\sin(x) \\\\ &=4x^3\sin(x)\cos(x)+x^4\cos^2(x)-x^4\sin^2(x) \end{aligned}$ In conclusion: $\dfrac{d}{dx}\left[ x^4\sin(x)\cos(x) \right]=4x^3\sin(x)\cos(x)+x^4\cos^2(x)-x^4\sin^2(x)$